Induction to prove formula

[erkme73]

After spending many hours lurking at this site gleaning valuable info on how to solve various problems, I've finally hit one I can't find - and am using this as my first post here (go easy )...

I'm taking a pre-calc class, and it assumes a solid understanding of college algebra - which I aced... 17 years ago. So many of the basic principles had to be (and continue to be) relearned. In this case, I think I'm clear on the induction method for proving equality for every positive integer. My dilemma is how to add/simplify two terms with the same variable exponent. For grins, I'm posting the whole problem.

1+2+2[sup:18v5mv90]2[/sup:18v5mv90]+2[sup:18v5mv90]3[/sup:18v5mv90]+… 2[sup:18v5mv90]n-1[/sup:18v5mv90] = 2[sup:18v5mv90]n[/sup:18v5mv90]-1

Step 1: Prove by substituting 1 for n:
2[sup:18v5mv90](1)-1[/sup:18v5mv90] = 2[sup:18v5mv90](1)[/sup:18v5mv90]-1
2[sup:18v5mv90]0[/sup:18v5mv90] = 2-1
1=1

Step 2: Assume n = k

Step 3: Show that n = k+1

1+2+2[sup:18v5mv90]2[/sup:18v5mv90]+2[sup:18v5mv90]3[/sup:18v5mv90]+… 2[sup:18v5mv90]k-1[/sup:18v5mv90] = 2[sup:18v5mv90]k[/sup:18v5mv90]-1

(2[sup:18v5mv90]k[/sup:18v5mv90]-1)+2[sup:18v5mv90](k+1)-1[/sup:18v5mv90] = 2[sup:18v5mv90](k+1)[/sup:18v5mv90]–1

2[sup:18v5mv90]k[/sup:18v5mv90]-1+2[sup:18v5mv90]k[/sup:18v5mv90] = 2[sup:18v5mv90]k+1[/sup:18v5mv90]–1

Any help showing how to get both sides of the equation to equal one another, or how to add 2[sup:18v5mv90]k[/sup:18v5mv90]+2[sup:18v5mv90]k[/sup:18v5mv90] would be appreciated!

 

[galactus]

\(\displaystyle 2^{k}+2^{k}=2\cdot 2^{k}=2^{k+1}\)

 

[erkme73]

\(\displaystyle 2^{k}+2^{k}=2\cdot 2^{k}=2^{k+1}\)

Thanks!

 

[JeffM]

Galactus as always gave you the right answer.

But I'd like to add a few points about the method of mathematical induction.

The proof has the form:
(1) If there is an integer k such that X is true for k, then X is true for (k + 1).
(2) X is true for integer m.
(3) Then X is true for any integer n \(\displaystyle \geq\\) m.

The type of proof is not limited to propositions about positive integers; it can be extended to cover any set of integers.

To avoid getting confused, I think it is helpful to use different letters in the proof. Use n for the general class of integers for which the proposition is to be proved true. Use k for the integer for which you assume that the proposition is true. After step 1 you have not shown the proposition to be true because you assumed its truth for k. What you have shown that IF it is true for k, it is true for the next, and so the next, and so the next, ad infinitum. But it is still a conditional proof. Step 2 proves the truth of the proposition for one integer. So now step 1 makes all the bigger integers fall like a row of dominoes. If you use n in the proof, it makes it seem that you are assuming what is to be proved.